3
$\begingroup$

In Special Relativity, when one talks about "the reference frame of a particle" it is quite clear what they mean.

First of all: one starts by setting up cartesian coordinates on the flat spacetime. Those are to be thought as coordinates relative to one observer at rest at the origin.

Then one considers the motion of the particle relative to this first observer we picked based on the cartesian coordinate system. This gives a parametrized path $x^\mu(\tau)$ for the particle.

The frame of the particle then actually means building a set of cartesian axes moving with the particle. So that when we write components $v^\mu$ relative to the particle's frame, we actually mean to compute the components relative to these moving axes. In that frame the particle is at rest, so that its motion is trivialy just $(\tau,0,0,0)$.

That is all fine. But in GR things are much more complicated. The main issue is: we don't have that initial global cartesian system related to that observer.

Actually all we have is: (i) the notion of charts, which are not necessarily tied down to any observer and (ii) the notion of observer as a pair $(\gamma,e)$ being $\gamma : \mathbb{R}\to M$ a timelike future-directed worldline and $e_\mu$ a set of four orthonormal vector fields along $\gamma$ such that $e_0 = \gamma'$.

Now if we want to talk about the reference frame of a particle what would that mean? I'm quite lost mainly because in SR the notion of observer was "global": we have a set of cartesian axes extending over all spacetime, which can register any event, anywhere and we tie this down to a single observer.

In GR an observer can be seen as a worldline together with axes being carried along it. The observer is local, in the sense that (i) to it there doesn't correspond any coordinate system, let a lone a global one, (ii) the observer can only assign components to tensors which exists in events in his worldline.

Considering all this: when we talk about "the reference frame of a certain particle" in GR, what do we mean, and how do we build it up in a mathematically precise way?

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Minkowski_space $\endgroup$ – MujjinGun Mar 27 '17 at 3:21
  • $\begingroup$ Thanks for trying to help @MujjinGun, but I'm afraid this really doesn't help much here. I'm talking about General Relativity where spacetime is a general four dimensional lorentzian manifold $(M,g)$, not Minkowski space. $\endgroup$ – user1620696 Mar 27 '17 at 3:46
  • $\begingroup$ At any event $E$ in spacetime, the tangent space $T_EM$ is a vector space, and a frame is an ordered basis for that vector space. $\endgroup$ – WillO Mar 27 '17 at 5:06
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/12221/2451 and links therein. $\endgroup$ – Qmechanic Nov 2 '17 at 19:00
2
$\begingroup$

I believe in GR you can only pass to the reference frame of the particle locally.

It is as simple as this: you always have a local set of frames with Minkowski metric (by equivalence principle); just pick one with the time coordinate labeling the (infinitesimal) proper time along the worldline.

You are absolutely right in that there is no single global reference frame of the particle.

$\endgroup$
0
$\begingroup$

My understanding of a "reference frame" or maybe more specific a rest frame in GR is closely related to the concept of the orthogonal projector. Let $\Sigma$ be a spacelike or timelike hypersurface of the manifold $\mathcal{M}$ and $\mathcal{T}_p(\mathcal{M})$ is the tangent space of $p\in\Sigma$, then $\mathcal{T}_p(\mathcal{M})$ can be orthogonally decomposed as $$\mathcal{T}_p(\mathcal{M})=\mathcal{T}_p(\Sigma)\oplus\mathrm{Vect}(\mathbf{n}), \tag{1}$$ where $\mathrm{Vect}(\mathbf{n})$ is the 1D subspace of $\mathcal{T}_p(\mathcal{M})$ generated by the normal vector $\mathbf{n}$ of $\Sigma$. The decomposition (1) has a corresponding projector $\vec{\gamma}$ \begin{align}\vec{\gamma}: \mathcal{T}_p(\mathcal{M}) &\rightarrow \mathcal{T}_p(\Sigma)\\ \mathbf{v}&\mapsto \mathbf{v} \pm (\mathbf{n} \cdot \mathbf{v})\mathbf{n},\tag{2}\end{align} where $+$ is for a timelike $\Sigma$ and - for a spacelike $\Sigma$. (2) can be expressed in the coordinate basis $\left\{\mathbf{e}_\alpha\right\}$ of $\mathcal{T}_p(\mathcal{M})$ as $$\gamma^\alpha_{~\beta}=\delta^\alpha_{~\beta} \pm n^\alpha n_\beta.\tag{3}$$ The projection operator induces metric tensor on $\Sigma$: $\gamma_{\alpha \beta}$ from which we could extract an adapted coordinate system. $\gamma_{\alpha \beta}$ is in general not flat.

Such projectors can be used to project into local rest frames, free falling/non-rotating "Eulerian"-observers and a like.

For example lets consider an observer $\mathcal{O}$ an a the timelike world line $\mathcal{L}$ in $\mathcal{M}$ with the future-directed unit vector $\mathbf{u}$ $$\mathbf{u}\cdot\mathbf{u}=-1$$ tangent to $\mathcal{L}$. $\mathbf{u}$ is the observers four-velocity. The subspace $\Sigma$ into which $\vec{\gamma}$ projects can be considered as the local rest space of the observer $\mathcal{O}$, since the velocity vanishes on $\Sigma$: $\vec{\gamma}(u)=0$. The component of $\mathbf{u}$ normal to $\Sigma$ is not necessarily trivial.

Finding such a frame does not require a specific (flat/Euclidean/...) frame to begin with as long as one can find/define a non-lightlike world line $\mathcal{L}$ one can construct an orthogonal projector $\vec{\gamma}$ and with it comes the notion of a local rest space/frame. The concept and definition is coordinate independent (eqs. (1) and (2)).

This would be some math related to the points already made by OP. I adapted the notation of [Eric Gourgoulhon, 2007, 3+1 Formalism and Bases of Numerical Relativity].

$\endgroup$
  • $\begingroup$ Thanks @M.J.Steil for the answer. So given one observer with four velocity $u$, the submanifold $\Sigma$ you define would be the usual $3$-dimensional space as seen by this observer? Is that the idea? $\endgroup$ – user1620696 Mar 28 '17 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.